Answer:
Explanation:
Given
initial velocity [tex]u=11.2\ m/s[/tex]
At maximum height velocity of ball is zero
using
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](0)^2 -(11.2)^2=2\times (-9.8)\times (s)[/tex]
[tex]s=\frac{11.2^2}{2\times 9.8}[/tex]
[tex]s=6.4[/tex]
time taken by the ball to reach the maximum height
[tex]v=u+at[/tex]
[tex]0=11.2-9.8\times t[/tex]
[tex]t=\frac{11.2}{9.8}[/tex]
[tex]t=1.142\ s[/tex]
At [tex]t=2\ s[/tex] height of ball is
[tex]h=ut+\\frac{1}{2}at^2[/tex]
[tex]h=11.2\times 2-\frac{1}{2}9.8\times (2)^2[/tex]
[tex]h=22.4-19.6[/tex]
[tex]h=2.8\ m[/tex]
i.e. ball is moving downward
height at [tex]v=5\ m/s[/tex]
[tex]v^2-u^2=2as[/tex]
[tex]s=\frac{25-125.4}{2\times (-9.8)}[/tex]
[tex]s=5.12\ m[/tex]
